ctg(x)<=1/√3 inequation

Given the inequality:
$$\cot{\left(x \right)} \leq \frac{1}{\sqrt{3}}$$
To solve this inequality, we must first solve the corresponding equation:
$$\cot{\left(x \right)} = \frac{1}{\sqrt{3}}$$
Solve:
Given the equation
$$\cot{\left(x \right)} = \frac{1}{\sqrt{3}}$$
transform
$$\cot{\left(x \right)} - 1 - \frac{\sqrt{3}}{3} = 0$$
$$\cot{\left(x \right)} - 1 - \frac{\sqrt{3}}{3} = 0$$
Do replacement
$$w = \cot{\left(x \right)}$$
Expand brackets in the left part

-1 + w - sqrt3/3 = 0

Move free summands (without w)
from left part to right part, we given:
$$w - \frac{\sqrt{3}}{3} = 1$$
Divide both parts of the equation by (w - sqrt(3)/3)/w

w = 1 / ((w - sqrt(3)/3)/w)

We get the answer: w = 1 + sqrt(3)/3
do backward replacement
$$\cot{\left(x \right)} = w$$
substitute w:
$$x_{1} = \frac{\pi}{3}$$
$$x_{1} = \frac{\pi}{3}$$
This roots
$$x_{1} = \frac{\pi}{3}$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} \leq x_{1}$$
For example, let's take the point
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + \frac{\pi}{3}$$
=
$$- \frac{1}{10} + \frac{\pi}{3}$$
substitute to the expression
$$\cot{\left(x \right)} \leq \frac{1}{\sqrt{3}}$$
$$\cot{\left(- \frac{1}{10} + \frac{\pi}{3} \right)} \leq \frac{1}{\sqrt{3}}$$

                  ___
   /1    pi\    \/ 3 
tan|-- + --| <= -----
   \10   6 /      3  
                

but

                  ___
   /1    pi\    \/ 3 
tan|-- + --| >= -----
   \10   6 /      3  
                

Then
$$x \leq \frac{\pi}{3}$$
no execute
the solution of our inequality is:
$$x \geq \frac{\pi}{3}$$

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